问题
解答题
(1)已知
(2)s=
求不超过S的最大整数[s]. |
答案
(1)猜想:
=1+
+1 n2 1 (n+1)2
.n2+n+1 n(n+1)
证明:
=1+
+1 n2 1 (n+1)2
=n2(n+1)2+(n+1)2+n2 n2(n+1)2
=n2(n+1)2+2n(n+1)+1 n(n+1)
;n2+n+1 n(n+1)
(2)∵
=1+n2+n+1 n(n+1)
-1 n
,1 n+1
∴s=1+1-
+1+1 2
-1 2
+1+1 3
-1 3
+…+1+1 4
-1 2005
=2005+1-1 2006
=20051 2006
,2005 2006
∴[s]=2005.